In the schematic below, a person is holding an object in her hand with the forea
ID: 1349018 • Letter: I
Question
In the schematic below, a person is holding an object in her hand with the forearm at a constant angle a with respect to the horizontal and with the muscle foreming an angle theta with the foream. Draw the free-body diagram on the schematic below. Include the music and joint reaction forces as the weight of the arm and of the object. Include all relevent angles and dimensions. The foream extends forward such that the angle alpha is held constant while yheta is allowed to vary. Determine the expression that describes the magnitude of musicle force F as a function of theta Assume that alpha = 2 inches, b = 7 inches , c = 14 inches, W = 10 Ibf, W = 4 Ibf, alpha = 15 Degree, and theta = 40 Degree. Solve for the muscle force F and the joint reaction forces F and F [ans: FM = 105.9 Ibf, FJX = 44.7 Ibf, FJY = 82.0Ibf ]Explanation / Answer
In equilibrium net torque = 0
torque due to weightof the object + torque due to weight of the arm = torque due to FM
torque due to weightof the object = Wobj *c*cosalfa
torque due to weightof the arm = Warm*b*coasalfa
torque due to FM = FM*a*costheta
Wobj *c*cosalfa + Warm*b*coasalfa = FM*a*costheta
FM = [ Wobj *c*cosalfa + Warm*b*coasalfa ] / a costheta
FM = cosalfa*[ Wobj *c + Warm*b ] / a costheta
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FM = cos15*((10*14)+(4*7))/(2*cos40)
FM = 105.9 lbf
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along horizantal
FM*sin(40-15) = Fjx
Fjx = 44.7
along vertical
FM*cos25 - Wobj - Warm = Fjy
Fjy = 82 lbf